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### Course: Trigonometry>Unit 1

Lesson 7: The reciprocal trigonometric ratios

# Using reciprocal trig ratios

Sal is given two sides in a right triangle and the cotangent of one of the angles, and he uses this information to find the missing side. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• Does Soh Cah Toa only work for right angle triangles?
• Sometimes in a triangle-based problem that doesn't have a right triangle, you can cut the triangle into a couple of pieces forming right triangles, and then you can figure out the dimensions and/or angles using soh cah toa and logic to get to your victory lap. This is especially true if you are given the altitude of a triangle and the length of one of the sides connecting the altitude to the base. With triangles, there are often several ways to get to the answer.
• I am trying to understand from trig 1.5 questions how these legs give missing information in the form of an equation. To me it seems that when it asks for some of these numbers they are already used in the trig ratios but they are not the answer, it is like I have to compare two forms of the ratio, like it will ask for the adj, and there are three equations with information you need to use, which video shows you how to use this properly?
• IE
BC = 48
AC = ?
AC is the opposite of B, Angle ABC
Sin(angleABC) = 7/25
cos(angleABC) = 24/25
tan(angleABC) = 7/24
7 is not the answer, although SOH CAH TOA says it is a ratio of the opposite side over the hypotenuse. Why is 7 not the answer? I tried the same on some other problems and they were correct, this is not straight forward in some way. Why? What operations or way do I need to approach this problem to get missing information? The hints make it look like an equality is needed, but why when it seems apparent sin would give a relevant side of the ratio of the same triangle, or can it be expressed as some other information like on another similar triangle?
• How do you use sin, cos, tan, csc, sec, and cot to find angles?
• Well, finding angles is another major use of trigonometry. For example, if you know that the ratio between two sides is 1 to 2, and the sides happen to be opposite and hypotenuse to an unknown angle A, then you can use the sine ratio to tell you that sin A = 1/2. When an angle is equal to 30 degrees, then the sine ratio is always 1/2. So then you use known ratios to tell you that the angle has to be 30 degrees.

Likewise, the tangent (tan) of 45 degrees is 1, which means the opposite side over the adjacent side is 1/1 which is equal to 1. So if you have a tangent ratio equal to 1, you know the angle is 45 degrees. There are many common angles that you will become familiar with if you continue your study of trigonometry. These common angles are simple ones like 0 degrees, 45 degrees, 30 degrees, 60 degrees, 90 degrees, and simple multiples and fractions of these. In more advanced math, the angles are given in radians, but these basic angles are used so often that you will know them by heart.

Beyond those simple ones, you can use tables of the trig ratios or calculators to translate complicated ratios such as sin B = 0.2113 to find out that angle B is about 12.2 degrees.
• What is the difference between cosine and arcsine then? I mean it looks the same to me...
• The cosine and the arcsine are not the same at all.
The arcsine is the inverse function of sine. The cosine is the sine of the complementary angle.

With angles measured in radians, here are what each of these functions is equal to:
arcsine x = -i ln(ix+ √(1-x²))
cos x = ½ [e^(ix)+e^(-ix)]

Thus, there is no particular connection between the arcsine and the cosine.
• This question peaked in my head at when the solution of the 6 trigonometric ratios are there. Is it possible that you could find all six trigonometric ratios if you just have the information about two trigonometric ratios or was this just an example that somehow had this?
• Actually, aside from worrying about the sign, you can find all six ratios given just one of them. Like Sal said at the end of the problem, you could have found the third side using the Pythagorean Theorem instead of the second ratio, or you could have used the inverse trig functions to find the measure of the angle and then applied all six functions to that angle. The only thing about the second ratio that was important to know is that the cotangent was positive -- if it was negative then some of the other ratios would have had to been negative as well.
• We're given a right triangle with two sides (5, square root of 41). Couldn't we use COS-1 of angle FDE to find the angle then SIN or TAN of angle FDE to find the length of the missing side? I don't think we even need the other information given to find the length of 'a'.
• You are correct, only one of the two given ratios is actually needed. You could also just use the Pythagorean Theorem to find the last side and go from there.
• I thought the reciprocal of a trig function like Sine was that function to the negative 1 power.
• A reciprocal is the inverse of the trig ratio. Sin^-1 is the inverse of sin itself. Good question.
• for a you could go to desmos graphing type 5^2+x^2=41 and check the x intercepts
• You could also do:
C^2 - B^2 = A^2
√41^2 - 5^2 = a^2

"a" and "A" are different. A is for the Pythagorean theorem, whereas "a" is the unknown variable you are solving for.
(1 vote)
• As angle theta increases from 0 degree to 90 degree , cos theta decreases from 1 to
0. why & how ?
• You can know this if you draw the cosine curve. Also there is a way to find that if you differentiate the curve piece wise. Also try to find the limit at 0 and pi/2 . Unless you draw and analyse the curve you will not able to find it.
(1 vote)
• Can you do a reciprocal of sin, cos, or tan on a calculator? I have do a question that ask: sec 10 degree

how do I do that on a calculator?